LeetCode-728. Self Dividing Numbers
2018-01-26 13:39
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Description
Example 1
Note
Solution 1(C++)
class Solution { public: vector<int> selfDividingNumbers(int left, int right) { vector<int> res; for(int i=left;i <= right;i++){ if(isDividingNumbers(i)){ res.push_back(i); } } return res; } bool isDividingNumbers(int &num){ vector<int> digits; int temp = num; while(temp != 0){ int i=temp % 10; digits.push_back(i); temp=temp / 10; } for(auto n:digits){ if(n !=0 && num%n == 0){ continue; } else { return false; } } return true; } };
Solution 2(C++)
class Solution { public: vector<int> selfDividingNumbers(int left, int right) { vector<int> res; int temp,j; for(int i=left;i <= right;i++){ temp=i; while(temp != 0){ j=temp % 10; if(j == 0 || i % j != 0) break; temp /= 10; } if(temp == 0) res.push_back(i); else continue; } return res; } };
算法分析
解法一与解法二并没有本质的区别,只是更改了一下程序的写法,不过这样更改之后,解法二确实比解法一缩短了一定的运算时间。程序分析
解法一与解法二的主要区别就是程序编写上的区别。这个就不啰嗦了。此外,可以复习一下获取整数的各个位数:while(temp != 0){ int i=temp % 10; digits.push_back(i); temp=temp / 10; }
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